$f(t) = 2t^{3}-2t^{2}+t-2(h(t))$ $h(t) = 3t^{2}+6t$ $ f(h(0)) = {?} $
Explanation: First, let's solve for the value of the inner function, $h(0)$ . Then we'll know what to plug into the outer function. $h(0) = 3(0^{2})+(6)(0)$ $h(0) = 0$ Now we know that $h(0) = 0$ . Let's solve for $f(h(0))$ , which is $f(0)$ $f(0) = 2(0^{3})-2(0^{2})-2(h(0))$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = 3(0^{2})+(6)(0)$ $h(0) = 0$ That means $f(0) = 2(0^{3})-2(0^{2})+(-2)(0)$ $f(0) = 0$